II. EL HISPANISMO EN CHECOSLOVAQUIA
2.5. Orígenes y evolución del hispanismo en la República Eslovaca
The simplest method for judging the economic viability of projects is the payback period method. It is a rough measure of the time it takes for an investment to pay for itself. More precisely, the payback period is the number of years it takes for an invest-ment to be recouped when the interest rate is assumed to be zero. When annual savings are constant, the payback period is usually calculated as follows:
Payback period First cost Annual savings
S P R E A D S H E E T S A V V Y
Spreadsheets can be indispensable when it comes to making present worth or annual worth compar-isons among projects. It is useful to spend a bit of time thinking about how to set up spreadsheet rows and columns before starting a comparison so you can do the analysis easily and can document your work. A careful setup will also allow you to check the individual components of your work as you go along.
The tables below demonstrate the use of the NPV and PMT Excel functions. The NPV func-tion takes the present worth of a series of cash flows. It assumes that the cash flows start at the end of period 1 and that the interest rate is i. In contrast to other Excel functions, NPV assumes that the cash flows are receipts, so the sign of cash flows does not need to be reversed. The PMT Excel func-tion computes the annual worth of a single cash flow that occurs at time 0 (assumed to be a disburse-ment) over N periods at an interest rate of i.
The tables show how to compute the present worth of cash flows in cells B5 through to B10 (B5:B10). The two methods shown are used to first compute the present worth of the individual cash flows and then to sum (cell C11), and then to use the NPV function (cell C14). The equivalent annual worth, in cells C12 and C15, is found by using either a compound interest factor (cell C12) or the PMT Excel function (cell C15).
In the top part of the table on the right, cell B2 contains the interest rate. It is good practice to put the interest rate in a cell that is referred to by other formulas rather than to enter it directly into the compound interest formula. The reason for this is twofold: First, the reader can see what the interest rate is for documentation purposes. Second, should you need to do a different computation with another interest rate, it is easier to change one cell than it is to re-enter the formulas.
For example, if a first cost of $20 000 yielded a return of $8000 per year, then the pay-back period would be 20 000/8000 2.5 years.
If the annual savings are not constant, we can calculate the payback period by deducting each year of savings from the first cost until the first cost is recovered. The number of years required to pay back the initial investment is the payback period. For example, suppose the saving from a $20 000 first cost is $5000 the first year, increas-ing by $1000 each year thereafter. By addincreas-ing the annual savincreas-ings one year at a time, we see that it would take just over three years to pay back the first cost (5000 6000 7000 8000 26 000). The payback period would then be stated as either four years (if we assume that the $8000 is received at the end of the fourth year) or 3.25 years (if we assume that the $8000 is received uniformly over the fourth year).
According to the payback period method of comparison, the project with the shorter payback period is the preferred investment. A company may have a policy of rejecting projects for which the payback period exceeds some preset number of years. The length of the maximum payback period depends on the type of project and the company’s financial situation. If the company expects a cash constraint in the near future, or if a project’s returns are highly uncertain after more than a few periods, the company will set a maximum payback period that is relatively short. As a common rule, a payback period of two years is often considered acceptable, while one of more than four years is unaccept-able. Accordingly, government grant programs often target projects with payback periods of between two and four years with the rationale that in this range the grant can justify economically feasible projects that a company with limited cash flow would otherwise be unwilling to undertake.
The payback period need not, and perhaps should not, be used as the sole criterion for evaluating projects. It is a rough method of comparison and possesses some glaring weak-nesses (as we shall discuss after Examples 4.8 and 4.9). Nevertheless, the payback period method can be used effectively as a preliminary filter. All projects with paybacks within the minimum would then be evaluated, using either rate of return methods (Chapter 5) or present/annual worth methods.
EXAMPLE 4.8
Elyse runs a second-hand book business out of her home where she advertises and sells the books over the internet. Her small business is becoming quite successful and she is considering purchasing an upgrade to her computer system that will give her more reliable uptime. The cost is $5000. She expects that the investment will bring about an annual savings of $2000, due to the fact that her system will no longer suffer long failures and thus she will be able to sell more books. What is the payback period on her investment, assuming that the savings accrue over the whole year?
Payback period 2.5 years_____________________쏋
EXAMPLE 4.9
Pizza-in-a-Hurry operates a pizza delivery service to its customers with two eight-year-old vehicles, both of which are large, consume a great deal of gas, and are starting to cost a lot to repair. The owner, Ray, is thinking of replacing one of the cars with a smaller, three-year-old car that his sister-in-law is selling for $8000. Ray figures he can save $3000,
$2000, and $1500 per year for the next three years and $1000 per year for the following two years by purchasing the smaller car. What is the payback period for this decision?
5000 2000 First cost
Annual savings
The payback period is the number of years of savings required to pay back the initial cost. After three years, $3000 $2000 $1500 $6500 has been paid back, and this amount is $7500 after four years and $8500 after five years. The payback period would be stated as five years if the savings are assumed to occur at the end of each year, or 4.5 years if the savings accrue continuously throughout the year. __________________쏋
The payback period method has four main advantages:
1. It is very easy to understand. One of the goals of engineering decision making is to communicate the reasons for a decision to managers or clients with a variety of backgrounds. The reasons behind the payback period and its conclusions are very easy to explain.
2. The payback period is very easy to calculate. It can usually be done without even using a calculator, so projects can be very quickly assessed.
3. It accounts for the need to recover capital quickly. Cash flow is almost always a problem for small to medium-sized companies. Even large companies sometimes can’t tie up their money in long-term projects.
4. The future is unknown. The future benefits from an investment may be estimated imprecisely. It may not make much sense to use precise methods like present worth on numbers that are imprecise to start with. A simple method like the payback period may be good enough for most purposes.
But the payback period method has three important disadvantages:
1. It discriminates against long-term projects. No houses or highways would ever be built if they had to pay themselves off in two years.
2. It ignores the effect of the timing of cash flows within the payback period. It disregards interest rates and takes no account of the time value of money.
(Occasionally, a discounted payback period is used to overcome this disadvantage.
See Close-Up 4.3.)
3. It ignores the expected service life. It disregards the benefits that accrue after the end of the payback period.
In a discounted payback period calculation, the present worth of each year’s savings is subtracted from the first cost until the first cost is diminished to zero. The number of years of savings required to do this is the discounted payback period. The main disadvan-tages of using a discounted payback period include the more complicated calculations and the need for an interest rate.
For instance, in Example 4.8, Elyse had an investment of $5000 recouped by annual savings of $2000. If interest were at 10 percent, the present worth of savings would be:
Year Present Worth Cumulative
Year 1 2000(P/F,10%,1) 2000(0.90909) 1818 1818 Year 2 2000(P/F,10%,2) 2000(0.82645) 1653 3471 Year 3 2000(P/F,10%,3) 2000(0.75131) 1503 4974 Year 4 2000(P/F,10%,4) 2000(0.68301) 1366 6340
Thus the discounted payback period is over 3 years, compared with 2.5 years calculated for the standard payback period.
CLOSE-UP 4.3 Discounted Payback Period
Example 4.10 illustrates how the payback period method can ignore future cash flows.
EXAMPLE 4.10
Self Defence Systems of Cape Town is going to upgrade its paper-shredding facility.
The company has a choice between two models. Model 007, with a first cost of R500 000 and a service life of seven years, would save R100 000 per year. Model MX, with a first cost of R100 000 and an expected service life of 20 years, would save R15 000 per year. If the company’s MARR is 8 percent, which model is the better buy?
Using payback period as the sole criterion:
Model 007: Payback period 500 000/100 000 5 years Model MX: Payback period 100 000/15 000 6.6 years It appears that the 007 model is better.
Using annual worth:
Model 007: AW 500 000(A/P,8%,7) 100 000 3965 Model MX: AW 100 000(A/P,8%,20) 15 000 4815 Here, Model MX is substantially better.
The difference in the results from the two comparison methods is that the payback period method has ignored the benefits of the models that occur after the models have paid themselves off. This is illustrated in Figure 4.4. For Model MX, about 14 years of benefits have been omitted, whereas for model 007, only two years of benefits have been left out. __________________________________________________________________쏋
1
Ignored
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0
1
Ignored
2 3 4 5 6 7 0
(a) Model 007
(b) Model MX
Figure 4.4 Flows Ignored by the Payback Period
R E V I E W P R O B L E M S
R E V I E W P R O B L E M 4 . 1
Tilson Dairies operates several cheese plants. The plants are all old and in need of reno-vation. Tilson’s engineers have developed plans to renovate all of them. Each project would have a positive present worth at the company’s MARR. Tilson has $3.5 million available to invest in these projects. The following facts about the potential renovation projects are available:
Project First Cost Present Worth
A: Renovate plant 1 $0.8 million $1.1 million B: Renovate plant 2 $1.2 million $1.7 million C: Renovate plant 3 $1.4 million $1.8 million D: Renovate plant 4 $2.0 million $2.7 million
Which projects should Tilson accept?
A N S W E R
Table 4.2 shows the possible mutually exclusive projects that Tilson can consider.
Table 4.2 Mutually Exclusive Projects for Tilson Dairies
Total First Total Present
Project Cost Worth Feasibility
Do nothing $0.0 million $0.0 million Feasible
A $0.8 million $1.1 million Feasible
B $1.2 million $1.7 million Feasible
C $1.4 million $1.8 million Feasible
D $2.0 million $2.7 million Feasible
A and B $2.0 million $2.8 million Feasible
A and C $2.2 million $2.9 million Feasible
A and D $2.8 million $3.8 million Feasible
B and C $2.6 million $3.5 million Feasible
B and D $3.2 million $4.4 million Feasible
C and D $3.4 million $4.5 million Feasible
A, B, and C $3.4 million $4.6 million Feasible A, B, and D $4.0 million $5.5 million Not feasible A, C, and D $4.2 million $5.6 million Not feasible B, C, and D $4.6 million $6.2 million Not feasible A, B, C, and D $5.4 million $7.3 million Not feasible
Tilson should accept projects A, B, and C. They have a combined present worth of
$4.6 million. Other feasible combinations that come close to using all available funds are B and D with a total present worth of $4.4 million, and C and D with a total present worth of $4.5 million.
Note that it is not necessary to consider explicitly the “leftovers” of the $3.5 million budget when comparing the present worths. The assumption is that any leftover part of the budget will be invested and provide interest at the MARR, resulting in a zero present worth for that part. Therefore, it is best to choose the combination of projects that has the largest total present worth and stays within the budget constraint.■
R E V I E W P R O B L E M 4 . 2
City engineers are considering two plans for municipal aqueduct tunnels. They are to decide between the two using an interest rate of 8 percent.
Plan A is a full-capacity tunnel that will meet the needs of the city forever. Its cost is
$3 000 000 now and $100 000 every 10 years for lining repairs.
Plan B involves building a half-capacity tunnel now and a second half-capacity tun-nel in 20 years, when the extra capacity will be needed. Each of the half-capacity tuntun-nels costs $2 000 000. Maintenance costs for each tunnel are $80 000 every 10 years. There is also an additional $15 000 per tunnel per year required to pay for extra pumping costs caused by greater friction in the smaller tunnels.
(a) Which alternative is preferred? Use a present worth comparison.
(b) Which alternative is preferred? Use an annual worth comparison.
A N S W E R
(a) Plan A: Full-Capacity Tunnel
First, the $100 000 paid at the end of 10 years can be thought of as a future amount that has an equivalent annuity.
AW 100 000(A/F,8%,10) PMT (0.8,10,100 000)
100 000(0.06903) 6903
Thus, at 8 percent interest, $100 000 every 10 years is equivalent to $6903 every year.
Since the tunnel will have (approximately) an infinite life, the present cost of the lining repairs can be found using the capitalized cost formula, giving a total cost of
PW(Plan A) 3 000 000 6903/0.08 3 086 288 Plan B: Half-Capacity Tunnels
For the first tunnel, the equivalent annuity for the maintenance and pumping costs is
AW 15 000 80 000(0.06903) 20 522
The present cost is then found with the capitalized cost formula, giving a total cost of
PW1 2 000 000 20 522/0.08 2 256 525
Now, for the second tunnel, basically the same calculation is used, except that the present worth calculated must be discounted by 20 years at 8 percent, since the second tunnel will be built 20 years in the future.
PW2 {2 000 000 [15 000 80 000(0.06903)]/0.08}(P/F,8%,20)
2 256 525(0.21455) ⬵484 137 PW(Plan B) PW1 PW2 2 740 662
Consequently, the two half-capacity aqueducts with a present worth of costs of $2 740 662 are economically preferable.
(b) Plan A: Full-Capacity Tunnel
First, the $100 000 paid at the end of 10 years can be thought of as a future amount that has an equivalent annuity of
AW 100 000(A/F,8%,10) 100 000(0.06903) 6903
Thus, at 8 percent interest, $100 000 every 10 years is equivalent to $6903 every year.
Since the tunnel will have (approximately) an infinite life, an annuity equivalent to the initial cost can be found using the capitalized cost formula, giving a total annual cost of
AW(Plan A) 3 000 000(0.08) 6903 246 903 Plan B: Half-Capacity Tunnels
For the first tunnel, the equivalent annuity for the maintenance and pumping costs is
AW 15 000 80 000(0.06903) ⬵20 522
The annual equivalent of the initial cost is then found with the capitalized cost formula, giving a total cost of
AW1 2 000 000(0.08) 20 522 180 522
Now, for the second tunnel, basically the same calculation is used, except that the annuity must be discounted by 20 years at 8 percent, since the second tunnel will be built 20 years in the future.
AW2 AW1(P/F,8%,20)
180 522(0.21455) ⬵38 731 AW(Plan B) AW1 AW2
180 522 38 731 219 253
Consequently, the two half-capacity aqueducts with an annual worth of costs of
$219 253 are economically preferable.■
R E V I E W P R O B L E M 4 . 3
Fernando Constantia, an engineer at Brandy River Vineyards, has a $100 000 budget for winery improvements. He has identified four mutually exclusive investments, all of five years’ duration, which have the cash flows shown in Table 4.3. For each alternative, he wants to determine the payback period and the present worth. For his recommendation report, he will order the alternatives from most preferred to least preferred in each case.
Brandy River uses an 8 percent MARR for such decisions.
A N S W E R
The payback period can be found by decrementing yearly. The payback periods for the alternatives are then
A: 4 years
B: 4.3125 or 5 years C: 2 years
D: 4.1 or 5 years
The order of the alternatives from most preferred to least preferred using the payback period method with yearly decrementing is: C, A, D, B. The present worth computations for each alternative are:
A: PW 100 000 25 000(P/A,8%,5)
100 000 25 000(3.9926)
185
B: PW 100 000 5000(P/F,8%,1) 10 000(P/F,8%,2)
20 000(P/F,8%,3) 40 000(P/F,8%,4) 80 000(P/F,8%,5)
100 000 5000(0.92593) 10 000(0.85734)
20 000(0.79383) 40 000(0.73503) 80 000(0.68059)
12 928
C: PW 100 000 50 000(P/F,8%,1) 50 000(P/F,8%,2)
10 000(P/F,8%,3)
100 000 50 000(0.92593) 50 000(0.85734)
10 000(0.79283)
2908
D: PW 100 000 1 000 000(P/F,8%,5)
100 000 1 000 000(0.68059)
580 590
The order of the alternatives from most preferred to least preferred using the present worth method is: D, B, A, C.■
Table 4.3 Cash Flows for Review Problem 4.3
Cash Flow at the End of Each Year
Alternative 0 1 2 3 4 5
A $100 000 $25 000 $25 000 $25 000 $25 000 $ 25 000
B 100 000 5000 10 000 20 000 40 000 80 000
C 100 000 50 000 50 000 10 000 0 0
D 100 000 0 0 0 0 1 000 000
S U M M A R Y
This chapter discussed relations among projects, and the present worth, annual worth, and payback period methods for evaluating projects. There are three classes of relations among projects: (1) independent, (2) mutually exclusive, and (3) related but not mutually exclusive. We then showed how the third class of projects, those that are related but not mutually exclusive, could be combined into sets of mutually exclusive projects. This enabled us to limit the discussion to the first two classes—independent and mutually exclusive. Independent projects are considered one at a time and are either accepted or rejected. Only the best of a set of mutually exclusive projects is chosen.
The present worth method compares projects on the basis of converting all cash flows for the project to a present worth. An independent project is acceptable if its present worth is greater than zero. The mutually exclusive project with the highest present worth should be taken. Projects with unequal lives must be compared by assuming that the projects are repeated or by specifying a study period. Annual worth is similar to present worth, except that the cash flows are converted to a uniform series.
The annual worth method may be more meaningful and does not require more compli-cated calculations when the projects have different service lives.
The payback period is a simple method that calculates the length of time it takes to pay back an initial investment. It is inaccurate but very easy to calculate.